of 1 57

of 2 57

Matter, Space, And Time

0.0 Six-fold Ratios………………………………………………………3

1.0 The Mass of a Proton……………………………………………. 4

2.0 Proton-Seconds, Proton Radius, Hydrocarbons……….5

3.0 The Natural Second……………………………………………….8

4.0 Theory For The Atoms……………………………………………11

5.0 The Principle of Least Action…………………………………..17

6.0 Rigorous Formulation of Proton Seconds…………………20

7.0 Theory of Compounds…………………………………………….24

8.0 The Nature of Time………………………………………………..27

Logos

9.0 Orbital Parameters……………………………………………….28

10.0 Logos…………………………………………………………………37

11.0 Conclusion……………………………………………………39

12.0 Graphic Equations……………………………………………….44

13.0 Sexagesimal…………………………………………….53!

14.0 The Cosmic Calendar……………………………………………56

We suggest it is worthwhile to explore six-fold symmetry by itself. It is shown that the radius of a

proton and its mass can be derived from a theory of space and time. To do this a theory for the

atom is achieved because the theory of the proton needs the unit of a second as a natural

duration which can be achieved from the atom that is made of these protons. A theory of the

atom needs to utilize Avogadro’s Number. This is done by introducing the concept of

intermediary mass. The concept of proton-seconds is introduced which results in a theory for

compounds, combinations of atoms (which are combinations of protons). This is done for doped

silicon and germanium which makes transistors and diodes enabling the engineering of logic

circuit technology. It is also shown that the unit of a second is not just a natural duration where

the atom is concerned, and hence the proton, but where the Earth-Moon-Sun orbital mechanics

are concerned.

The recurrent six-fold symmetry follows through to the orbital parameters of celestial bodies

which we show in the case of orbital velocity is 6 for all bodies, and that centripetal acceleration

is six squared, and the orbital radii one. These values have to be translated into units that we can

understand by introducing the calibration of the measuring stick used. These integer values for

orbital parameters are converted by units used in a matrix transformation with the measuring

stick matrix multiplied by the integer values of 1, 6, 6^2. The orbital parameters of the orbiting

bodies are perfect, and the approximate answer in units we can apply is approximate not

because of the motions of bodies in a gravitational field, but because of the nature of the man-

made measuring stick. We can call this discovery Logos. The integer orbital parameters are

arrived at by actually looking at the patterns of the moon and sun in the sky over time, which is

how we divided the calendar since ancient times resulting in the duration of a second from a

sexagesimal counting system (base 60). We see the formation of the calendar from these

observations as the moon and sun appear in the sky it would turn out are connected to the

actually their actual physical nature. This prompts us to discuss the sexagesimal and its origins,

and to introduce a cosmic calendar in sync with earth events.

of 3 57

0.0 Six-fold Ratios We suggest there is an aspect of Nature founded on six-fold symmetry, the

example of which we are interested in here is The Periodic Table of the Elements, because it has

18 groups which we can define by carbon, C. This because we have the following scenario:

Equations 1

And, we pull out the 2 and the 3 and write (Fig. 1)

, ,

Where , such that

which is given by

In general

Equations 2: ,

, ,

3 + 3 + 3 = 9

2 + 2 + 2 = 6

3 ⋅ 6 = 18

2 ⋅ 9 = 18

2 ⋅ 3 = 6

2cos

= 2

2cos

5 + 1

2cos

= 3

5 + 1

= Φ =

a = b + c

ϕ =

−

− 1 = 0

f (n) = 2cos

n = 4,5,6

= 45

∘

π /5 = 36

∘

= 30

∘

Fig. 1 Dividing line in

golden mean,

of 4 57

1.0 The Mass of a Proton Gravity is a property of space measured by the universal constant

of gravity, G:

Equation 1.1

Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons

and neutrons) we will suggest given by:

Equation 1.2.

Which describes mass per meter over time, which is:

Equation 1.3:

It must be adjusted by the fine structure constant . It is my guess the factor should be which

is 18,769.:

Equation 1.4.

Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a

hypersphere. Thus we consider the surface area of a proton, :

Equation 1.5.

We take the square root to get meters:

Equation 1.6.

We multiply that with the value we have in equation 1.4:

Equation 1.7.

We find that the mass of a proton realizes if we divide this by six seconds:

Equation 1.8.

That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:

Equation 1.9.

Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the

skeletons of organic matter, the so-called hydrocarbons:

G = 6.674E − 11N

= 6.674E − 11

⋅ kg

kg ⋅ s

1.82E − 16

kg ⋅ s

(1.82E − 16kg ⋅ s/m)(18,769) = 3.416E − 12kg ⋅ s /m)

= 4πr

= 8.72E − 30m

= 2.953E − 15m

(2.953E − 15m)(3.416E − 12

kg ⋅ s

) = 1.009E − 26kg ⋅ s

1E − 26

6secon ds

= 1.67E − 27kg

1E − 26kg ⋅ s

1secon d

= 1E − 26kg = 6protons = 6m

of 5 57

Equation 1.10.

Equation 1.11.

If we divide 1E-26kgs by something greater than 6 seconds we get fractional protons. The rest of

the elements in the periodic table occur for dividing by something less that 1-second. It seems

the duration of a second is natural. If it is, since it was formed by a calendar based on

reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon

orbital mechanics. I find it is, that:

That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic

energy of the earth is about 1 second (about 1.2 seconds).

All of the computations thus far are shown on the next page…

2.0 Proton-Seconds, Proton Radius, Hydrocarbons We show carbon, the core element

of life is six-fold symmetric with hydrogen in terms of the natural constants that characterize

space, time, and matter:

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

We find one second gives six protons which is carbon:

Equation 2.1

We find six seconds gives 1 proton is hydrogen:

Equation 2.2

1E − 26kg ⋅ s

6secon ds

1E − 26kg ⋅ s

1secon d

K E

moo n

K E

earth

(Ear th Da y) ≈ 1second

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G : 6.67408 × 10

−11

c : 299,792,459m /s

α : 1/137

h 4π r

= 6proton ⋅ secon d s = carbon(C )

of 6 57

For time t greater than 6 seconds we have fractional protons. For t<6 we the have other

elements.

Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by

the mass of a proton. But these masses can be considered to cancel and leave pure number. We

make a program that looks for close to whole number solutions so we can create a table of values

for problem solving.

By what value would you like to increment?: 0.25

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

24.1199 protons 0.250000 seconds 0.119904 decpart

12.0600 protons 0.500000 seconds 0.059952 decpart

8.0400 protons 0.750000 seconds 0.039968 decpart

6.0300 protons 1.000000 seconds 0.029976 decpart

4.0200 protons 1.500000 seconds 0.019984 decpart

3.0150 protons 2.000000 seconds 0.014988 decpart

2.1927 protons 2.750000 seconds 0.192718 decpart

2.0100 protons 3.000000 seconds 0.009992 decpart

1.2060 protons 5.000000 seconds 0.205995 decpart

1.1486 protons 5.250000 seconds 0.148567 decpart

1.0964 protons 5.500000 seconds 0.096359 decpart

1.0487 protons 5.750000 seconds 0.048691 decpart

1.0050 protons 6.000000 seconds 0.004996 decpart

0.2487 protons 24.250000 seconds 0.248659 decpart

0.2461 protons 24.500000 seconds 0.246121 decpart

0.2436 protons 24.750000 seconds 0.243635 decpart

A very interesting thing here is looking at the values generated by the program, the smallest

integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds

produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of

the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons

the backbones of biological chemistry. Here is the code for the program:

#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

int n;

float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792459,protons[100],r=0.833E-15;

{

printf("By what value would you like to increment?: ");

scanf("%f", &increment);

h 4π r

= 1proton ⋅ 6seconds = hydr ogen(H )

h 4π r

of 7 57

printf("How many values would you like to calculate for t in equation 1 (no more than 100?):

");

scanf("%i", &n);

}

while (n>=101);

{

for (int i=0; i<n;i++)

{

protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));

int intpart=(int)protons[i];

float decpart=protons[i]-intpart;

t=t+increment;

if (decpart<0.25)

{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);

}}}}

We need an interpretation of equation 2.1, which was

Matter is that which has inertia. This means it resists change in position with a force applied to

it. The more of it, the more it resists a force. We understand this from experience, but what is

matter that it has inertia? In this analogy we are suggesting a proton is a three dimensional

bubble embedded in a two dimensional plane. As such there has to be a normal vector holding

the higher dimensional sphere in a lower dimensional space. Thus if we apply a force to to the

cross-section of the sphere in the plane there should be a force countering it proportional to the

normal holding it in a lower dimensional universe. It is actually a 4-dimensional hypersphere

whose cross-section is a sphere. This counter force would be experienced as inertia. (Fig. 2)

Since plank’s constant h is a measure of energy over time where space and time are concerned it

must play a role. Of course the radius of a proton plays a role since squared and multiplied by

it is the surface area of our proton embedded in space. The gravitational constant is force

produced per kilogram over a distance, thus it is a measure of how the surrounding space has an

effect on the proton giving it inertia. The speed of light c has to play a role because it is the

velocity at which events are separated through time. The mass of a proton has to play a role

because it is a measurement of inertia itself. And alas the fine structure constant describes the

degree to which these factors have an effect. We see the inertia then in equation 6 is six protons

over 1 second, by dimensional analysis.

h 4π r

= 6proton ⋅ secon d s = carbon(C )

4π

Fig. 2

of 8 57

We see the radius of a proton is given by carbon (1 second):

Equation 2.3.

The experimental radius of a proton is:

Equation 2.4.

The fine structure constant squared is the ratio of the potential energy of an electron in the first

circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of

light squared:

Equation 2.5.

3.0 The Natural Second In that we get one second for carbon and 6 seconds for hydrogen

very nearly even, that is

Eq. 3.1

It is suggested that the second is a natural unit. If it is, since it comes from designing a calendar

that reconciles the phases of the moon with the Earth year (12 moons per year, approximately) it

is suggested the unit of a second should be in the Earth-Moon-Sun orbital mechanics. The

translational kinetic energy of the moon and earth are:

To find the translational kinetic energy of the moon:

Distance from earth: 3.85E8m

Orbital period:

T=27.32 days=2.36E6seconds

v=1.025E3m/s~1000m/s

Mass: 7.34767E22kg

Use

E=3.67E28 Joules

To find the translational kinetic energy of the earth:

Distance from Sun: 1AU=1.496E11m

Orbital period: 1 year=3.1558E7 seconds

v=2.9785E4m/s, Earth mass: 5.9722E24kg

4πh

= 8.288587 × 10

−16

m = 0.829f m

= 0.833

0.014f m

⋅

h 4π r

= 1.004996352secon d s

K . E . Moon = 3.67E 28J

K . E . Ear th = 2.649E 33j

2π (3.85E8m) = 2.419E 9m

E =

2π (1.496E11m) = 9.399E11m

of 9 57

E=2.649E33 Joules

Earth day=(24)(60)(60)=86,400 seconds.

It turns out:

Eq . 3.2

Where the Lunar Month can be as much as 31 days and is based on the lunar orbital period

(27.32 days). We have

Eq. 3.3

Essentially we have formed a Planck constant, h, for the moon by multiplying its kinetic energy

over the time for the period of its orbit:

If we let the lunar month cancel with moon’s orbital period, we have:

Eq. 3.4

Since

is units of mass divided by we can let it cancel with , the mass of a proton, and write:

Eq. 3.5

That is:

1secon d ≈

(K Eof Moon)(LunarOrbital P er iod )

LunarMo nth

EarthDay

(K Eof E ar th)

31d a ys

(1Ear th Da y)

= 31 ≈ π

= 31.006

h = (3.67E 28J )(2.36E6s) = 8.6612E 34J ⋅ s

8.6612E 34J ⋅ s

K Eof E ar th

= 32.696secon ds

h 4π r

≈ 6proton s ⋅

K Eof Moon

K Eof E ar th

⋅ Ear th Da y

1sec

⋅

h 4π r

≈ 6 ⋅

K Eof Moon

K Eof E ar th

⋅ Ear th Da y = (6)1.2secon d s

6proton s

⋅

h 4π r

= 1.00secon d

of 10 57

is phenomenal because It allows multiplication between degrees and seconds to output our

fundamental ratios ( ). We see in the following wave:

, ,

Where t=1 second is carbon yielding:

And, t=6 seconds is hydrogen yielding:

In so far as

relates carbon=1second to the Earth-Moon-Sun orbital mechanics and to the radius of a proton

through six-fold symmetry:

We have the following around which all our structure is based

, ,

2, 3, . . .

A = A

cos(θ t)

= 1

θ = 30

∘

,60

∘

,45

∘

A(60

∘

) = cos(60

∘

⋅ 1s) = 0.5

A(30

∘

) = cos(30

∘

⋅ 1s) = 3/2

A(45

∘

) = cos(45

∘

⋅ 1s) = 2/2

A(60

∘

) = cos(60

∘

⋅ 6s) = 1

A(30

∘

) = cos(30

∘

⋅ 6s) = − 1

A(45

∘

) = cos(45

∘

⋅ 6s) = 0

h 4π r

≈ 6 ⋅

K Eof Moon

K Eof E ar th

⋅ Ear th Da y ≈ (6)1secon d

h 4π r

= 6

f (n) = 2cos

n = 2,3,4,5.6.,...

f (2) = 2

f (5) = Φ = 1/ϕ

f (6) = 3

of 11 57

4.0 Theory For The Atoms We have determined the mass and radius of the proton, but still

need to derive the second to open it up to other realms and I find this can be done by deriving a

constant k by theorizing the atom, something made up of protons, neutrons and electrons. But I

have derived such a constant from Giordano’s relationship.

We now formulate what I call Giordano’s Relationship: Warren Giordano writes in his paper

The Fine Structure Constant And The Gravitational Constant: Keys To The Substance Of The

Fabric Of Space, March 21, 2019:

In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic

equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’

is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational constant

numerically, but neglecting any units.

Let’s do that

(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js

And it works, G is:

G=6.67408E-11 N(m2/kg2)

Let us reformulate this as:

Equation 4.1

Where and H=1 gram/atom

Because for hydrogen 1 proton is molar mass 1 gram, for carbon 6 protons is 6 grams and so on

for 6E23 atoms per gram. Thus,…

Since grams and atom cancel we can work in grams even though our equations are in kilograms.

Let us not write H, since formally it is grams per mole of hydrogen but write

We have:

Or,…

Equation 4.2

Where

1 + α

(1 + α)

H = 6.0003

⋅ s

= Avaga dr o′ s − Nu m ber = 6.02E 23a tom s /gra m

H = 6.02E 23

atom s

gr a m

⋅

1gr a m

atom

= 6.02E 23

ℍ = 1

gr a m

atom

(1 + α)

⋅ N

ℍ = 6.0003kg

⋅

h(1 + α)N

ℍ = 6Gx

of 12 57

Equation 4.3

Let us say we were to consider Any Element say carbon . Then in general

Equation 4.4

We have

and

Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a

molar mass of 12. We have

12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus

And it follows that

We see in general since the atomic number Z is the number of protons in an atom that in general

this holds for all elements because

And,

Therefore we always have:

Equation 4.5

This works nicely because we formulated molar mass nicely; we said element one (hydrogen)

which is one proton and one electron has one gram for a mole of atoms. Historically this was

done because we chose carbon (element six) to have 12 grams per mole, and determined what

the mole was such that it would hold. The reason this works is that hydrogen is one proton and

has no neutrons, but carbon has twelve neutrons but since hydrogen doesn’t have any neutrons,

and the neutron has the same mass as the proton, and our theory makes use only of protons (in

this instance of its formulation) equation 4.3

x = 1.00kg

𝔼

ℂ

(1 + α)

⋅ N

𝔼 = 6.0003kg

⋅

ℂ =

6gr a m s

6proton s

6(6E 23pr oton s)

6gr a m s

ℂ = 6E 23

(1 + α)

⋅ N

ℂ = 6.0003kg

⋅

𝔼

Z ⋅ 6E 23pr oton s

Z ⋅ gram s

𝔼 =

Z ⋅ gram s

Z ⋅ pr oton s

⋅ 𝔼 = 6E 23

x = 1.00kg

of 13 57

Comes out to have x equal to 1.00 (nearly) even. It is at this moment that we point out, because

it is important, that in equation 4.5

is not molar mass, and that is a variable determined by ; it is the number of a mole of

atoms multiplied by the number of protons in . The reason we point this out, though it may

already be clear, is we wish to find the physical theory behind it. That is we need to find the

physical explanation for equation 4.4

It is the integer 6 to 3 ten thousandths. Which classifies it as interesting because since it is in

kilograms, seconds, and meters, it may mean these units of measurement have some kind of a

meaning. We can in fact write it:

We know that

The fine structure constant squared is the ratio of the potential energy of an electron in the first

circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of

light squared. To begin our search for the meaning of equation 1.4 we convert x, the factor of

1.00 to astronomical units, years, and solar masses, as these are connected to the orbit

of earth as it relates to the sun. We have:

We can now write

Eq 4.6.

This unit of AU/year is very interesting. It is not , which would be the Earth’s orbital

velocity, but is a velocity given by the earth orbital radius to its orbital period, which is quantum

mechanical in nature. It relates to earth as as a state, as we have with atoms, a number. We

⋅ 𝔼 = 6E 23

𝔼

(1 + α)

⋅ N

𝔼 = 6.0003kg

⋅

(1 + α)

⋅ N

𝔼 = 6.000kg

⋅

(1.98847E 30)

⊙

1.4959787E11m

⋅

year

3.154E 7s

1.8754341E64

⊙

⋅ year

(1 + α)

⋅ N

𝔼

year

= 8.2172E 32M

⊙

2π AU/year

of 14 57

multiply both sides by and we have earth velocity on the left and the units stay the same on

the right. But what we will do is return to the form in kg-m-s and leave it as an equation but put

in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:

Eq. 4.7

This brings up an interesting question: while we have masses characteristic of the microcosmos

like protons, and masses characteristic of the macrocosmos, like the minimum mass for a star to

become a neutron star as opposed to a white dwarf after she novas (The Chandrasekhar limit)

which is 1.44 solar masses, we do not have a characteristic mass of the intermediary world where

we exist, a truck weighs several tons and tennis ball maybe around a hundred grams. To find

that mass let us take the geometric mean between the mass of a proton and the mass of 1.44

solar masses. We could take the average, or the harmonic mean, but the geometric mean is the

squaring of the proportions, it is the side of a square with the area equal to the area of the

rectangle with these proportions as its sides. We have:

We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .

We have the intermediary mass is:

Eq. 4.8

All we really need to do now is divide 4.7 by 4.8 and we get an even number that is the six of our

six-fold symmetry.

Eq. 4.9

The six of our six-fold symmetry.

We have something very interesting here. We have

This is:

Equation 4.10

Where k is a constant, given

Equation 4.11

We can take the velocity of earth as being 30,000 m/s by rounding it. We have

4π

(1 + α)

⋅ N

𝔼 ⋅ v

= 422.787kg

⊙

= 1.98847E 30kg

= 1.67262E − 27kg

= (2.8634E 30)(1.67262E − 27) = 69.205kg

(1 + α)

⋅ N

𝔼 ⋅ v

= 6.1092 ≈ 6

69.205kg

6kg

= 6

k v

= 6

k =

800

30,000

800

= 37

of 15 57

Using , we write

37.5 = 6.123734357

⋅

h 4π r

= 1.004996352secon d s

k v

= 6

h 4π r

Gck v

= 1proton ⋅ secon d

(K . E . Moon)(Ear th Da y)

(K . E . Ear th)

≈ 1secon d

h 4π r

Gck v

(K . E . Ear th)

(K . E . Moon)(Ear th Da y)

= 1proton

of 16 57

Eq. 4.12

The primordial element from which all others were made.

We can explicitly write the constant k:

Equation 4.13

It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf

will not have its gravity overcome the degeneracy pressure and collapse. The non-relativistic

equation is:

Equation 4.14

Let us approximate 0.77 with 3/4. Since we have our constant

And

Then

Equation 4.15

Since our constant k in terms the Chandrasekhar limit is

Equation 4.16

h 4π r

(K . E . Ear th)

(K . E . Moon)(Ear th Da y)

= 6proton s

h 4π r

Gck v

(K . E . Ear th)

(K . E . Moon)(Ear th Da y)

= ℍ

ℍ = hydrogen

k =

(1 + α)

⋅ N

𝔼

M ≤ 0.77

ℏ

= 1.41 ⊙

k =

1 + α

⋅ N

𝔼

= Mm

ℏ

1/2

ℏ = h /2π

k =

8π

(1 + α) ⋅ N

𝔼

of 17 57

5.0 The Principle of Least Action We consider two paths, one with velocity c in one

medium, the other with velocity v in another. In order to go from one point two another over

two paths, the refraction is such that the sine of the angle of incidence equals the sine of the

angle of refraction.

We have the two paths are travelled in a time t:

This is the mysterious Nature of reality, the path of least time is taken. This falls under the

general heading The Principle of Least Action, attributed to firstly the French Natural

Philosopher and mathematician Louis Maupertuis of the early eighteenth century. I say

mysterious, because as it is said in physics, for something to know the path of least action and

take it, it is as if it has explored first all paths between A and B to know which one would be the

path of least action. Everything in physics comes to this principle. It is called a principle, not a

theory, law, or rule. Yet it seems to be the way Nature behaves, and it is mysterious. Richard

Feynman applied it to quantum mechanics, probably because the mysterious Planck’s constant

that governs quantum mechanics is in Joule-seconds, energy over time, and this is the terms in

which action is formulated mathematically.

In our scenario here we regard matter, the proton in particular, as the cross-section of a

hypersphere. Our two mediums are hyperspace and space and the least action principle applies

in the same mathematical form. This is abstract cosmology, that really is the underlying

mathematics is common to all systems, that in effect they are manifestations of one another.

We have

We make the approximation

So that

We are saying t=6 seconds is the proton, and r is the radius of a proton r=0.833E-15m. Thus

The radius of a hydrogen atom is .

sinα

sinβ

t = r

c + v

c v

c v − v = c

c v = c

v =

0.833E − 15m

= 1.389E − 16m /s

= 1.2E − 10m

of 18 57

And we have our one second as a natural constant with respect to the atom. We see it occurs at

t/2 which is at half the radius of a hydrogen atom. We want to deal with that size or around it;

our velocity comes from it and it can be thought of as a proton drift velocity akin to the electron

drift velocity in a wire that gives rise to an electrical current. We find if we derive the equation

that represents these computations, we use t not t/2 which is not because of the

approximations, made, the right answer is achieved. If we are to derive the equation we have

Where,…

, ,

Radius of hydrogen atom

Remember our constant k equation 4.16:

t =

1.2E − 10m

1.389E − 16m /s

= 863,930,.8855s

= 431,965.4428s

= (773.5m /s)(431,965.4428s) = 334,125,270m

334,125,270m

299,792,459m /s

= 1.11452193secon d s

v =

t = 6s =

h 4π r

= α

h 4π r

v =

= α

h 4π

= 1.2E − 10m

t =

= R

⋅

h 4π

⋅

8π

⋅

1 + α

⋅

𝔼

≈

⋅

8π

⋅

𝔼

3 2

⋅

Gπ

⋅

𝔼

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.2secon d s

of 19 57

of 20 57

6.0 Rigorous Formulation of Proton Seconds

We can actually formulate this differently than we have. We had

But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2 are

lower and upper limits in an integral, then we have:

Equation 6.1

This Equation is the generalized equation we can use for solving problems.

Essentially we can rigorously formulate the notion of proton-seconds by considering

Equation 6.2

Is protons-seconds squared where current density is and ( can also be

). We say

Equation 6.3

Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a

number of protons. It is

Equation 6.4

Dividing Equation 6.2 through by t:

Equation 6.5

Which is proton-seconds. Dividing through by t again:

h 4π r

= 6proton s

h 4π r

= 1proton

h 4π r

∫

dt = ℕ

∫

qd t = t

∬

ρ(x, y, z)d x d y



J = ρ



ρ = Q /m

Q /m

Q =

∫

ρdV

ℕ =

h 4π r

∫

= t

∬

ρ(x, y, z)d x d y

of 21 57

Equation 6.6

We see that if where and then J is I/m2 (current per square meter)

is analogous to amperes per per square meter which are coulombs per second through a surface.

Thus we are looking at a number of protons per second through a surface. Thus we write:

Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to

protons in the output from our program and 1.0 seconds is carbon (C). We see we have the

following theorem:

Equation 6.7

So as an example,…

Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second

which is carbon and we have protons per square meter.

h 4π r

∫

= proton s



J = ρ



ρ = Q /m

v = m /s

h 4π r

∫

= 6

∫

1.0

0.5

−2

dt = − 6(1 − 2) = 6

h 4π r

∫

∬



J ⋅ d



h 4π r

∫

1.0

0.5

∬



J ⋅ d



S = − 3

(

1 −

0.25

)

= 9

proton s

secon d



J(x, y, z) = (0,0, J ) = − J



S = d xd y



J ⋅ d



S = (0,0, J ) ⋅ (0,0, d xd y) = − Jd x d y

of 22 57

We are now equip to do computations in proton-seconds. We use equation 7.6 from two to

three, the smallest prime numbers that multiply to make six-fold symmetry in our hexagonal

proton that we found described its radius (My feeling is we introduce the factor of 2 because

carbon is 6 protons +6 neutrons and 2 times 6 is twelve):

But we already know this is the radius of a proton in fm, it is as if fm is a natural length, as well

as the second we have been using. 2 seconds in our program is 3 protons (Lithium Li) and 3

seconds is two protons is helium, He:

By what value would you like to increment?: 0.01

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

86.1425 protons 0.070000 seconds 0.142517 decpart

50.2498 protons 0.120000 seconds 0.249805 decpart

43.0713 protons 0.140000 seconds 0.071259 decpart

40.1998 protons 0.150000 seconds 0.199841 decpart

30.1499 protons 0.200000 seconds 0.149876 decpart

26.2173 protons 0.230000 seconds 0.217283 decpart

25.1249 protons 0.240000 seconds 0.124893 decpart

24.1199 protons 0.250000 seconds 0.119900 decpart

23.1922 protons 0.260000 seconds 0.192213 decpart

20.0999 protons 0.300000 seconds 0.099920 decpart

17.2285 protons 0.350000 seconds 0.228504 decpart

15.0749 protons 0.400000 seconds 0.074944 decpart

14.0232 protons 0.430000 seconds 0.023204 decpart

13.1086 protons 0.460000 seconds 0.108647 decpart

12.0600 protons 0.500000 seconds 0.059957 decpart

11.1666 protons 0.540000 seconds 0.166627 decpart

h 4π r

∫

= proton s

h 4π r

∫

= 0.833pr oton s /secon d

of 23 57

10.2203 protons 0.590000 seconds 0.220302 decpart

10.0500 protons 0.600000 seconds 0.049964 decpart

9.1363 protons 0.660000 seconds 0.136332 decpart

8.1486 protons 0.740000 seconds 0.148621 decpart

8.0400 protons 0.750000 seconds 0.039971 decpart

7.1785 protons 0.839999 seconds 0.178546 decpart

7.0941 protons 0.849999 seconds 0.094093 decpart

7.0116 protons 0.859999 seconds 0.011603 decpart

6.2165 protons 0.969999 seconds 0.216474 decpart

6.1530 protons 0.979999 seconds 0.153040 decpart

6.0909 protons 0.989999 seconds 0.090889 decpart

By what value would you like to increment?: 0.25

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

24.1199 protons 0.250000 seconds 0.119904 decpart

12.0600 protons 0.500000 seconds 0.059952 decpart

8.0400 protons 0.750000 seconds 0.039968 decpart

6.0300 protons 1.000000 seconds 0.029976 decpart

4.0200 protons 1.500000 seconds 0.019984 decpart

3.0150 protons 2.000000 seconds 0.014988 decpart

2.1927 protons 2.750000 seconds 0.192718 decpart

2.0100 protons 3.000000 seconds 0.009992 decpart

1.2060 protons 5.000000 seconds 0.205995 decpart

1.1486 protons 5.250000 seconds 0.148567 decpart

1.0964 protons 5.500000 seconds 0.096359 decpart

1.0487 protons 5.750000 seconds 0.048691 decpart

1.0050 protons 6.000000 seconds 0.004996 decpart

0.2487 protons 24.250000 seconds 0.248659 decpart

0.2461 protons 24.500000 seconds 0.246121 decpart

0.2436 protons 24.750000 seconds 0.243635 decpart

Thus we write

Equation 6.8.

Thus we suggest the radius of a proton of a proton (0.833protons) is the proton radius .

Now we integrate from Aluminum is 13 protons=0.372 seconds to phosphorus is 15 protons =

0.396 seconds which is to integrate across silicon:

By what value would you like to increment?: 0.006

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

251.2490 protons 0.024000 seconds 0.248978 decpart

59.1174 protons 0.102000 seconds 0.117416 decpart

50.2498 protons 0.120000 seconds 0.249805 decpart

40.1998 protons 0.150000 seconds 0.199844 decpart

37.2221 protons 0.162000 seconds 0.222080 decpart

27.1621 protons 0.222000 seconds 0.162058 decpart

25.1249 protons 0.240000 seconds 0.124907 decpart

20.0999 protons 0.300000 seconds 0.099916 decpart

17.0338 protons 0.354000 seconds 0.033823 decpart

16.2096 protons 0.372000 seconds 0.209604 decpart

h 4π r

∫

= 0.833pr oton s /secon d = r

proton

of 24 57

15.2272 protons 0.396000 seconds 0.227202 decpart

14.1549 protons 0.426000 seconds 0.154862 decpart

13.2236 protons 0.456000 seconds 0.223620 decpart

13.0519 protons 0.462000 seconds 0.051885 decpart

12.1084 protons 0.498000 seconds 0.108375 decpart

11.1666 protons 0.540000 seconds 0.166615 decpart

11.0439 protons 0.546000 seconds 0.043906 decpart

10.1515 protons 0.594000 seconds 0.151471 decpart

Equation 6.9.

Silicon can be doped with phosphorus to make negative (n-type) silicon that semi-conducts thus

enabling the construction of logic circuits that you can use to make computing machines. But

this must be joined with positive (p-type) silicon which usually uses boron, but boron is in the

same group as aluminum, just above it. This results in a theory for AI elements as mathematical

constructs, that we will go into now.

7.0 Theory of Compounds

Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the

elements. As you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and

germanium are in group 14 meaning they have 4 valence electrons and want 4 for more to attain

noble gas electron configuration. If we dope Si with B from group 13 it gets three of the four

electrons and thus has a deficiency becoming positive type silicon and thus conducts. If we dope

the Si with P from group 15 it has an extra electron and thus conducts as well. If we join the two

types of silicon we have a semiconductor for making diodes and transistors from which we can

make logic circuits for AI.

As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the

right of Si but doping agent B is not directly to the left, aluminum Al is. This becomes important.

I call (As-Ga) the differential across Ge, and (P-Al) the differential across Si and call Al a dummy

in the differential because boron B is actually used to make positive type silicon.

That the AI elements make a three by three matrix they can be organized with the letter E with

subscripts that tell what element it is and it properties, I have done this:

h 4π r

∫

= − 6(6.377 − 7.22) = 5proton s /secon d = boron

of 25 57

Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the

first row and has 4 valence electrons making it carbon (C). I believe that the AI elements can be

organized in a 3 by 3 matrix makes them pivotal to structure in the Universe because we live in

three dimensional space so the mechanics of the realm we experience are described by such a

matrix, for example the cross product. Hence this paper where I show AI and biological life are

mathematical constructs and described in terms of one another.

We see, if we include the two biological elements in the matrix (E14) and and (E15) which are

carbon and nitrogen respectively, there is every reason to proceed with this paper if the idea is to

show not only are the AI elements and biological elements mathematical constructs, they are

described in terms of one another. We see this because the first row is ( B, C, N) and these

happen to be the only elements that are not core AI elements in the matrix, except boron (B)

which is out of place, and aluminum (Al) as we will see if a dummy representative makes for a

mathematical construct, the harmonic mean. Which means we have proved our case because the

first row if we take the cross product between the second and third rows are, its respective unit

vectors for the components meaning they describe them!

The Computation



A = (Al, Si, P)



B = (G a, G e, As)



A ×



B =



Al Si P

Ga Ge A s

= (Si ⋅ A s − P ⋅ Ge)



B + (P ⋅ G a − Al ⋅ A s)



C + (Al ⋅ G e − Si ⋅ Ga)





A ×



B = − 145



B + 138



C + 1.3924



A = 26.98

+ 28.09

+ 30.97

= 50g/m ol

B = 69.72

+ 72.64

+ 74.92

= 126g/m ol



A ⋅



B = A Bcosθ

cosθ =

6241

6300

= 0.99

θ = 8

∘



A ×



B = A Bsi nθ = (50)(126)sin8

∘

= 877.79

of 26 57

And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological

elements C and N being the unit vectors are multiplied by the AI elements, meaning they

describe them. But we have to ask; Why does the first row have boron in it which is not a core

biological element, but is a core AI element? The answer is that boron is the one AI element that

is out of place, that is, aluminum is in its place. But we see this has a dynamic function.

The Dynamic Function

The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon

(Si) and germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic

(As) have an asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C)

and as such have 4 valence electrons. Thus to have positive type silicon and germanium, they

need doping agents from group 13 (three valence electrons) like boron and gallium, and to have

negative type silicon and germanium they need doping agents from group 15 like phosphorus

and arsenic. But where gallium and arsenic are in the same period as germanium, boron is in a

different period than silicon (period 2) while phosphorus is not (period 3). Thus aluminum (Al)

is in boron’s place. This results in an interesting equation.

Equation 7.1.

The differential across germanium crossed with silicon plus the differential across silicon

crossed with germanium normalized by the product between silicon and germanium is equal to

the boron divided by the average between the germanium and the silicon. The equation has

nearly 100% accuracy (note: using an older value for Ge here, it is now 72.64 but that makes the

equation have a higher accuracy):

Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic

mean between the semiconductor elements (by molar mass):

Equation 7.2

877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol

Si(A s − G a) + Ge(P − Al )

SiG e

Ge + Si

28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)

(28.09)(72.61)

2(10.81)

(72.61 + 28.09)

0.213658912 = 0.21469712

0.213658912

0.21469712

= 0.995

(A s − G a) +

(P − Al ) =

2SiGe

Si + Ge

of 27 57

8.0 The Nature of Time We need offer an interpretation of time, as well. (Fig. 3)

In The Time Machine, a science fiction story by HG Wells, the time traveller describes time as

physical distance, the direction through which the universe is falling at the speed of light, c.

Thus, not only when we move through space do we travel through a distance at a velocity v, but

we travel through a distance t at a velocity c. If we draw the picture and account for that distance

and velocity as well, we arrive at time dilation as given by relativity theory. He writes:

I think that at the time none of us quite believed in the Time Machine. The fact is, the Time

Traveller was one of those men who are too clever to be believed: you never felt that you saw all

around him; you always suspected some subtle reserve, some ingenuity in ambush, behind his

lucid frankness.

x = vt

+ t

= d

Fig. 3

of 28 57

Since we don’t experience our motion through time as we fall through it with the universe, we

change the sign in . And have the relativistic equation for time dilation:

9.0 Orbital Parameters The recurrent six-fold symmetry follows through to the orbital

parameters of celestial bodies which we show in the case of orbital velocity is 6 for all bodies,

and that centripetal acceleration is six squared, and the orbital radii one. These values have to be

translated into units that we can understand by introducing the calibration of the measuring

stick used. These integer values for orbital parameters are converted by units used in a matrix

transformation with the measuring stick matrix multiplied by the integer values of 1, 6, 6^2. The

orbital parameters of the orbiting bodies are perfect, and the approximate answer in units we

can apply is approximate not because of the motions of bodies in a gravitational field, but

because of the nature of the man-made measuring stick. The integer orbital parameters are

arrived at by actually looking at the patterns of the moon and sun in the sky over time, which is

how we divided the calendar since ancient times resulting in the duration of a second from a

sexagesimal counting system (base 60). We see the formation of the calendar from these

observations as the moon and sun appear in the sky it would turn out are connected to the

actually their actual physical nature. This prompts us to discuss the sexagesimal and its origins,

and to introduce a cosmic calendar in sync with earth events.

Earth rotates through 15 degrees in one hour:

The distance then that the earth equatorial surface rotates through is where r is the earth

radius and theta is in radians.

Now we look at how many degrees through which the earth rotates in 1 minute:

= 1

(

+ 1

)

= t

1 +

1 −

360

= 15

s = r θ

s = (6,378.14)(0.2618ra d ) = 1669.8k m /hr

24(60) = 1440min

of 29 57

For seconds…

Now let’s look at the same for the distance the earth moves around the Sun in an hour, a minute,

and a second as well…

And finally…

We have the following table:

360

1440

= 0.25

∘

= 0.0043633r a d

s = (6,378.14)(0.00436) = 27.83k m /min

24(60)(60) = 86,400s

360

∘

86,400

= 0.004167

∘

= 0.00007272r a d

s = (6,378,14)(0.00007272) = 0.464k m /s

(365.25)(24) = 8766hr /yr

360

8766

= 0.041

∘

= 0.00071678r a d

(149,598,000)(0.000716768) = 107,227k m /hr

(365.25)(24)(60) = 525960

360

525960

= = 0.000684463

∘

= 0.000011946r a d

s = (149,598,000)(0.000011946) = 1,787.1k m /min

(365.25)(24)(60)(60) = 31557600

360

31557600

= 0.000011408

∘

= 0.000000199r a d

(149,598,000)(0.000000199) = 29.786k m /sec

of 30 57

As we can see I am in good agreement with Martin Zombek, Handbook of Space Astronomy and

Astrophysics which provides data. Notice 27.83 km/min in earth rotation is approximately the

29.786 km/sec in earth orbit. That is a valuable clue. Now let us consider

Earth: 1 AU=149,598,000km (average earth-sun separation).

The earth rotates through about one degree a day.

Earth Orbit:

Earth Rotation: r=6,378.14km

Moon: sidereal month =27.83 days, r=405,400 km

The points to be made in this exploration

1. So the earth goes through 1 degree a day and the moon 1 kilometer per second.

2. Earth rotates through 27.83 km/min at its equatorial surface and orbits through 29.786km/s

around the sun. These two are close to the same.

3. The synodic month is 29.53 days approximately equals the 29.786 km/s the Earth moves

around the sun. The sidereal month is 27.83 days is the 27.83 km/min through which the

Earth rotates at its surface.

360

365.25

= 0.9856

deg

d a y

≈ 1

deg

d a y

360

∘

= 2π ra di a n s = 6.283

ra d

year

0.9856

∘

= 0.017202424

ra d

d a y

(149,598,000)(0.017202424) = 2,573,448.201

k m

d a y

360

∘

1d a y

= 6.283

ra d

d a y

s = r θ = (6,378.14)(6.283) = 40,075.0

k m

d a y

(27.83)(24) = 667.92

h ours

m onth

360

667.92

= 0.529

∘

= 0.0094

ra d

(405,400k m)(0.0094ra d /hr) = 3,810.76k m /hr

(3,810.75k m /hr)(hr /60min) = 63.5k m /min = 1.0585k m /s

of 31 57

With that information we have this mystery of sexagesimal in the earth-moon-sun orbital

parameters, solved. We see we can make the following equation in table form:

Consider minute times day:

Thus,..

The circle is divided into 360 units, each unit (each degree) is the distance the earth moves

around the Sun in a day, where a day is one turn of the earth on its its axis, and as such there are

360 such turns in the time it takes the earth to go around the sun approximately (365.25 days).

We have:

1 astronomical unit (AU) is the distance of the earth from the sun on the average, and is always

close to that because its orbit is approximately circular. We have

This is approximately the diameter of the Earth orbit. We define our variables:

Earth orbits:

Earth rotates:

Earth orbits:

Moon orbits:

Earth completes a 360 degree orbit yields:

orbit rotat ion orbit m oon

29.786k m minute 1d a y k ilom eter

secon d 27.83k m degree secon d

min ⋅ d a y

deg

⋅

k m

(min)(d a y) = 60(24 ⋅ 60 ⋅ 60) = 864,000sec

min ⋅ d a y

deg

⋅

k m

86,400s

deg

⋅

k m

= 86,400 ⋅

k m

deg

86,400 ⋅

k m

deg

⋅ 360

∘

= 311,040,000k m

311,040,000k m

149,598,000k m /AU

= 2.079AU

= 29.786

k m

27.83k m

min

27.83k m

min

⋅

min

60sec

= 0.4638

k m

sec

1d a y

deg

(24)(60)(60)

deg

= 14,400

sec

deg

1k m

sec

⋅ θ

⋅ v

⋅

360

∘

1AU

= 2π

⋅ θ

⋅ v

⋅ r

of 32 57

Where on the right it is in radians and is the radius of the Earth’s orbit. We have

This is (0.00618)360=2.225

0.00618 is

Where is the inverse of the golden ratio. .

We have:

Equation 8.1

Our base ten counting is defined

is defined

, such that

which is given by

= 1AU

(29.786)(14,400)(1k m /s)

(0.4638)(149,598,000)

360 = 2.225AU

100

ϕ = 0.618 =

100

360

∘

= 2AU

= 1,10

= 100,...

5 + 1

= Φ =

a = b + c

ϕ =

−

− 1 = 0

of 33 57

Thus since the diameter of the Earth orbit is

Then its radius is

Since we measure time with the Earth orbital period and that period is given by Kepler as

Then

Equation 8.2

Is approximately one year. In this section we set out two show the historical development of the

second by dividing up the motions of the Earth, Moon, and the apparent motion of the Sun into

units of 24, and 60 result in the solar system’s size as based around the Earth orbit giving us

Where this duration of a second is in the atomic world in carbon (six protons) the core element

of life which we found was:

And that the duration of a second was as we showed in the orbital mechanics of the Earth and

moon as such

Since we have established a connection between the microcosmos and the macrocosmos we

would do well to introduce the units of AU (astronomical unit), year, solar masses. Thus we want

to know the universal gravitational constant in these units:

100

360

∘

= 2AU

⋅

100

360

∘

= 1AU

= a

T =

10 ⋅

⋅ 360

∘

100

360

∘

= 2AU

6proton s

⋅

h 4π r

= 1.00secon d

(K . E . Moon)

(K . E . Ear th)

(Ear th Da y) ≈ 1second

of 34 57

For a year we have

(365.25)(24)(60)(60)=31557600 seconds

And for an AU

1 AU=1.496E11m

We can immediately put this to work. In order for the earth to stay in orbit its centripetal force

must equal the gravitational force. Its orbital velocity must be given by:

Since in our theory we present R as:

Then

Equation 8.3

Which evaluates:

G = 6.67408E − 11

kg ⋅ s

G = 6.67408E − 11

kg ⋅ s

⋅

3.3479E 33m

⋅

1.98847E 30kg

⊙

⋅

9.9477E14s

year

G = 39.433

⊙

⋅ year

G ≈ 40

⊙

⋅ year

⊙

= 40

year

⊙

4π

= 0.99885 ≈ 1

v =

⋅

100

360

∘

= 1AU

v =

of 35 57

Converting this to meters per second:

Which should be about right. The orbital velocity is given in the data tables as about 30,000m/s

is an average over a varying velocity due to the Earth’s slightly elliptical orbit.

Since we have

For the earth, we show now it is true for the moon as well, and indeed it would hold for any

system circular, or approximately circular. For the Sun and its planets

If you use astronomical units and Earth years. In general we use

Which has the constant of proportionality

If m is small compared with M, you can write

Since we want to work with Earth and the Moon m is not so small compared to M. So,…

And these do round neatly to one-exponent like this. The sidereal lunar month is

(27.83)(24)(60)(60)=2,404,513 seconds. Thus,…

v =

(40)(1)

0.618

= 6

year

⋅

year

3.156E 7

⋅

1.496E11m

= 28,441

⋅

100

360

∘

= 1AU

= a

= G

M + m

4π

= G

M + m

4π

= G

4π

= (6.674E − 11)

(

5.9722E 24kg + 7.347731E 22kg

4π

)

(6.674E − 11)

(

6E 24kg

4π

)

4E14

4π

1E14

= 1E13

of 36 57

The tables give

Which is an average for the lunar orbit in its slightly elliptical orbit.

We need to find G in terms of Lunar Units (LU) as opposed to Astronomical Units (AU), and in

terms of the lunar month as one instead of the earth year. We see G has the same value we had

for solar-earth units:

We only need to convert to m, kg, s by using their relationships with LU, LM, and . That is,

for the orbital velocity is always six:

Where is the orbital velocity of the moon, and is the orbital velocity of the earth. The

difference comes when converting the system from lunar units in the case of the first to kg/m/s

and from astronomical units in the case of the second to kg/m/s. For example the first is:

Which is correct, the tables give that the lunar orbital velocity is 1000 m/s. This factor of six is

the factor recurrent throughout our work in this paper, like

= (1E13)(2,404,512)

= 5.78E 25

a = 3.8668E8m

a = 3.833E8m

G = 6.674E − 11

kg ⋅ s

⋅

(3.8668E8m)

⋅

5.9722E 24kg

⊕

⋅

(2,404,512s)

L M

40.896 ⋅

⊕

⋅ L M

⊕

(40)(1)

= 6

(40)(1)

= 6

3.8668E8m

2,404,512s

= 964.886m /s

h 4π r

= 6proton ⋅ secon ds = carbon(C )

of 37 57

10.0 Logos

To find logos, is to find Nature not just as number, but as dimensionless whole number, because

fractions have inherent in them irrational numbers, which are unending decimal expressions

that at some point must be rounded to so many figures to put them to use. Logos then is the

language of Nature that transcends conundrum because it would be utilized as relationship

between point, plane, and line as opposed to units. We have found that for the orbital velocity:

Equation 10.1

Where is the earth orbital radius. This results in the orbital velocity for a orbiting body is

Equation 10.2.

Which is logos because it is number, which is 6. The orbital velocity of a body is always 6

because:

Where is the orbital velocity of the moon, and is the orbital velocity of the earth, That is G

is always 40, and M is always 1 providing the orbit is circular. Let us show this for Venus. Its

orbital distance VU (Venus units) is 1.082E11meters. It orbital period is 1.94E7s is the Venus

year (VY).

Thus we have

We can then express all orbital velocities as 6, but to find there values in a formal system of

units, we need to convert from these natural units to something like kg/m/s. Thus logos

translated into a language we understand can best be done in a square array, or as a matrix

transformation. We have

⋅

100

360

∘

= 1AU

v =

(40)(1)

= 6

(40)(1)

= 6

G = 6.674E − 11

kg ⋅ s

⋅

(1.082E11m)

⋅

1.989E 30kg

⊙

⋅

(1.94E 7s)

V Y

39.44

V U

⊙

V Y

≈ 40

(40)(1)

= 6

of 38 57

You will find this gives , ,

All of which are correct within the variations of these velocities in their deviations from a

perfectly circular orbit. Logos is:

Equations 10.3

, such that

which is given by

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E11m

31,557,600s

1.082E11m

19,400,000s

= 964.886m /s

= 28,443m /s

= 33,463.9m /s

(

)

⋅

100

360

∘

= 1LU

⋅

100

360

∘

= 1AU

⋅

100

360

∘

= 1V U

v =

5 + 1

= Φ =

a = b + c

ϕ =

−

− 1 = 0

of 39 57

11.0 Conclusion To say that logos is:

Is to say that for the solar system Kepler’s Law of Planetary Motion holds:

That in general

Because as such for Earth we have

For the Moon we have

For Venus we have

Making

, ,

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E11m

31,557,600s

1.082E11m

19,400,000s

= a

= G

M + m

4π

G = 6.67408E − 11

kg ⋅ s

⋅

3.3479E 33m

⋅

1.98847E 30kg

⊙

⋅

9.9477E14s

year

G ≈ 40

⊙

⋅ year

G = 6.674E − 11

kg ⋅ s

⋅

(3.8668E8m)

⋅

5.9722E 24kg

⊕

⋅

(2,404,512s)

L M

40.896 ⋅

⊕

⋅ L M

G = 6.674E − 11

kg ⋅ s

⋅

V U

(1.082E11m)

⋅

1.989E 30kg

⊙

⋅

(1.94E 7s)

V Y

39.44

V U

⊙

V Y

≈ 40

(40)(1)

= 6

(40)(1)

= 6

(40)(1)

= 6

of 40 57

This is derived from

Which is to say

Is logos because

such that

which is given by

Which is the relationship between point, plane, and line.

v =

⋅

100

360

∘

= 1AU

⋅

100

360

∘

= 1LU

⋅

100

360

∘

= 1AU

⋅

100

360

∘

= 1V U

v =

5 + 1

= Φ =

a = b + c

ϕ =

−

− 1 = 0

of 41 57

From everything we have said G, the gravitational constant is about 40

We can do this for any planet and get G is approximately 40. We found the orbital velocity of any

planets is 6. This is true because as we have shown, the orbital distance of any planet is

This gives since

That orbital velocities are for the moon, earth, and venus

But what does this mean? It means since

And, v=6, that where the orbital velocities of the planets are 6, their distances from the sun are

all one.

This gives:

G = 6.674E − 11

kg ⋅ s

⋅

(Ear th Dista n ceMeters)

⋅

Sol arMassKilogr a m s

⊙

⋅

(Ear thOrbita lPer iod Secon ds)

year

⋅

100

360

∘

= 1AU

v =

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E11m

31,557,600s

1.082E11m

19,400,000s

R =

≈ 1

(

1 1 1

)

3.8668E8m

1.496E11m

1.082E11m

of 42 57

But where six is

Six is also

And one is

And one is also

Where one second is given by Earth-moon orbital mechanics

12.0 Summary What we have done here is capture the six-fold logos across the whole

spectrum:

Which are the biological life skeletons the hydrocarbons. Shown that carbon the core element of

life is one second predicts the radius of the proton:

That the second itself

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E11m

31,557,600s

1.082E11m

19,400,000s

1secon d

h 4π r

= 6proton s = carbon(C )

(

1 1 1

)

3.8668E8m

1.496E11m

1.082E11m

6secon ds

h 4π r

= 1proton = h ydr ogen(H )

(K . E . Moon)

(K . E . Ear th)

(Ear th Da y) ≈ 1second

h 4π r

= 6proton ⋅ secon ds = carbon(C )

h 4π r

= 1proton ⋅ 6secon d = hydrogen(H )

4πh

= 8.288587 × 10

−16

m = 0.829f m

of 43 57

Is a natural unit

We have brought this together:

Have described all orbital velocities as 6:

Have brought in chemistries’ Avogadro’s number with a constant k that that determines the

integer 6 in terms of the earth orbital velocity above:

But where six is

Six is also

And one is

⋅

h 4π r

= 1.004996352secon d s

(K . E . Moon)

(K . E . Ear th)

(Ear th Da y) ≈ 1second

h 4π r

≈ 6proton s ⋅

K Eof Moon

K Eof E ar th

⋅ Ear th Da y

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E11m

31,557,600s

1.082E11m

19,400,000s

k =

8π

(1 + α) ⋅ N

𝔼

k v

= 6

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E11m

31,557,600s

1.082E11m

19,400,000s

1secon d

h 4π r

= 6proton s = carbon(C )

(

1 1 1

)

3.8668E8m

1.496E11m

1.082E11m

of 44 57

And one is also

Where one second is given by Earth-moon orbital mechanics

12.0 Graphic Equations We can say visually since the orbital velocity of a planet or a moon is

6, that it is the perimeter of a unit regular hexagon in logos:

Since the radius R of the orbit in logos is 1, then its visual logos is

the radius R of this regular hexagon because the radius of a

regular hexagon is equal to its side (s). We have

, ,

The six equilateral triangles that make up the regular hexagon are

computed here:

The height of each equilateral triangle is . Since then and since

is a regular hexagon, is a square because :

6secon ds

h 4π r

= 1proton = h ydr ogen(H )

(K . E . Moon)

(K . E . Ear th)

(Ear th Da y) ≈ 1second

R =

⋅

100

360

∘

v =

G = 40

= cos

3 = 2cos

= 45

∘

of 45 57

And, is a regular pentagon which contains is the golden ratio is :

These are the regular polygons that tessellate (tile a plane without leaving gaps). They tessellate

in the plane, except for the regular pentagon, which tessellates closed, it encloses a volume, and

is 1 of 4 other solids that do the same which are made up of our tessellating regular polygons, the

equilateral triangle, the regular hexagon, and the square:

The area of a regular pentagon is given by

A =

5(5 + 2 5s

of 46 57

For s=1 this is . This gives the surface are of a regular dodecagon is:

Continue to next page…

A = 1.72 ≈ 1.732 = 3

of 47 57

Actually let us denote the surface area of the dodecagon as

And reserve A for the area of a regular pentagon, which we

can write either

Or,…

Much in the same way that in vector calculus

Are both surface integrals. We want to use surface area of a pentagon as

Even though it shows a dodecahedron, because it speaks of the connection of the regular

pentagon to the dodecahedron. We want denote surface areas and volumes

= Volume of Dodecahedron

∫

d xd y =

∬

d xd y

of 48 57

With actual images and shapes of these objects, because logos is to have a deeper understanding

of Nature, and this allows us to work with the ideas in our mind in connection to the equations.

We have seen that the orbit of any planet or moon can be taken as 1:

And that 1 is astronomical unit (AU), moon unit (MU), Venus unit (VU), Mars unit (MU),…and

so on. The gravitational constant G can be taken as 40 for every planet, but in units that

correspond to the planet (or body) in particular:

Where M is the mass of the body orbited, like the Sun for the Earth, the Earth for the moon, and

is the planet’s orbital radius, it’s orbital period and these in any system of units, for

example in kg/m/s for the Earth

For the moon is the orbital radius in meters from the Earth, is the Earth mass, the mass

of the body it orbits, is its orbital period. The orbital velocity of any planet can be taken as 6

And must be translated from this the logos to a system of units we understand, like kg/m/s. In

general the components of logos are given by

n=1,2,3,4,5,6,…

In that

, , where , ,…

We have orbital velocity is hexaperimeter, and orbital radius is hexaradius:

⋅

100

360

∘

= 1

G = 6.674E − 11

⋅ s

⋅

= 1.496E1m = 1AU = Ear thOrbit

⊙

= 1.98847E 30kg = Sol arMa ss

= 31,557,600s = Ear thYear

⊕

v =

(40)(1)

0.618

= 6

f (n) = 2cos

2cos

= 3

2cos

= Φ

ϕ =

2cos

= 2

of 49 57

We are going to suggest that where orbital velocity is associated with the regular hexagon, the

orbital radius is associated with the square, and the centripetal acceleration is associated with

the dodecahedron and pentagon. We have the mass of the earth in solar masses is

Which is correct. Where

Thus we have for the orbit radii, orbital velocities, and centripetal accelerations of an orbiting

body

⊕

5.9722E 24

1.98847E 30

= 3.00E − 6M

⊙

= 0.00003M

⊙

= 36

year

⋅

100

360

∘

= 1AU

2cos(36

∘

) = 2cos

360

∘

2ϕ

360

∘

⋅

cos

−1

2ϕ

= 1

360

∘

= cos

−

2ϕ

= 36

∘

24 cos(36

∘

) ≈ 20

(

1 1 1

)

3.8668E8m

1.496E11m

1.082E11m

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E11m

31,557,600s

1.082E11m

19,400,000s

(

)

3.8668E8m

(2,404,512s)

1.496E11m

(31,557,600s)

1.082E11m

(19,400,000s)

of 50 57

Thus we say

That,…

Is orbital radius, is:

Is orbital velocity, is:

Is centripetal acceleration, is:

Where we have used

The image of an icosahedron as the area of a regular hexagon, because a

regular hexagon is its shadow.

(

1 1 1

)

3.8668E8m

1.496E11m

1.082E11m

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E11m

31,557,600s

1.082E11m

19,400,000s

(

)

3.8668E8m

(2,404,512s)

1.496E11m

(31,557,600s)

1.082E11m

(19,400,000s)

of 51 57

Aside from our logos formulation of the planets being useful theoretically, it is useful for orbital

mechanics computations whether it be of planets or spacecraft because it uses multiplication

of integer values like six for orbital velocity that can be converted to units we understand in

matrix form, which is at the heart of writing computer programs in their linear algebra

applications which to the computer are arrays. Further since we translate from logos to units,

by dividing out large numbers (like astronomical units and seconds in a year) in such a way that

they are not overly large numbers, yet large enough that they can be written as integers without

losing much accuracy and there are many simple programs on the internet that multiply

matrices with integer elements. This one comes from programmingsimpliﬁed.com:!

#include <stdio.h>

int main(int argc, const char * argv[]) {

int m,n,p,q,c,d,k,sum=0;

int first[10][10],second[10][10],multiply[10][10];

printf("Enter number of rows and columns of matrix 1: \n");

scanf("%d %d",&m,&n);

printf("Enter elements of matrix 1: \n");

for(c=0;c<m;c++)

for(d=0;d<n;d++)

scanf("%d",&first[c][d]);

printf("Enter number of rows and column of matrix 2: \n");

scanf("%d %d", &p,&q);

if(n!=p)

printf("The multiplication can't be done.");

else

{

printf("Enter the elements of matrix 2: \n");

for(c=0;c<p;c++)

for(d=0;d<q;d++)

scanf("%d", &second[c][d]);

for(c=0;c<m;c++){

for(d=0;d<q;d++){

for(k=0;k<p;k++){

sum=sum+first[c][k]*second[k][d];

}

multiply[c][d]=sum;

sum=0;

}

printf("Product of the matrices is: \n");

for(c=0;c<m;c++){

for(d=0;d<q;d++)

printf("%d\t", multiply[c][d]);

printf("\n");

}

return 0;

}

of 52 57

We need to divide out our data for the planets and the moon, which are 9 elements, which works

great because our program works only up to 3 by 3 matrices. We have

{3.8668E8m}{2,404,512s^2}=161=moon

{1.496E11m}{31,557,600s^2}=4,740=earth

{1.082E11m}{19,400,000s^2}=5,577=venus

You can do the rest of the planets, but we will just do these and put zeros in for the remaining 6

elements. Running the program we have

Enter number of rows and columns of matrix 1:

Enter elements of matrix 1:

161

4740

5577

Enter number of rows and column of matrix 2:

Enter the elements of matrix 2:

Product of the matrices is:

966 0 0

28440 0 0

33462 0 0

These answers are in meters per second.These answers are fairly accurate but

can be made more accurate by adjust the orbital velocity 6 according to

orbital pecularities, which would be a subject in itself that is outside the

scope of our work here.

of 53 57

13.0 Sexagesimal

My book Abstract Cosmology could have easily been called The Book of Six, it was

Buckminster Fuller who said “Nature employs 60 degree coordination”. Sixty degrees are the

degrees of an equilateral triangle. The triangle is the structure that encloses an area with the

least amount of sides. Buckminster Fuller said, systems of triangles are the the only inherently

stable patterns. The regular hexagon, a six-sides polygon tessellates as equilateral, 60 degree,

inherently stable equilateral triangles. It was the scientist Shubnikov who said among the living

organisms the pattern with which we most frequently meet is ﬁve-fold symmetry. It is well

known that the physical, like snowﬂakes, are six-fold symmetry. Thus we should not be

surprised that our Abstract Cosmology is founded on six-fold symmetry. Two and three are the

smallest prime numbers and their product is six. Two times six is twelve, the number most

evenly divisible by whole numbers for its size (it is a so-called abundant number: divisible

evenly by 1,2,3,4,6, their sum is 16 which is greater than twelve itself and, three times six is

eighteen, the cyclical Nature of the periodic table (18 groups). !

At the dawn of civilization, the Sumerians who settled down from wandering, gathering, and

hunting, to invent agriculture and and build ceramic homes, developed the ﬁrst mathematics,

and it was this sexagesimal (base 60) that passed on to the Babylonians, then ended up with

the Ancient Greeks, who divided not just the hour into 60 minutes of time and the minute into

60 seconds of time, but who divided the sky into hours, minutes, and seconds of arc which

was measured in time counted by the rotation of the Earth, and its orbital period around the

Sun, and the orbital period of the Moon around the Earth.!

We are suggesting nature is founded on six-fold symmetry, six which is constructed by

multiplying together the ﬁrst two prime numbers, 2 and 3. Two factorial is two, three factorial is

6. Two times six is twelve, the number of months in a year, approximated by our moon’s

approximately 12 orbits around the earth in the time it takes the earth to go around the sun

once. There are four weeks in a lunar month, and!

Is the number of seconds in the 24 hour Earth day. The duration of one second comes from

dividing up time and the sky like this in base 60, that came from the Sumerians, Babylonians,

and Ancient Greeks and it is believed they did this because 60 is evenly divisible by!

1,2,3,4,5,6,…12,15,20,30,60,…!

The moon orbits the Earth in approximately 30 rotations of the Earth, hence the 30 day month.

January has 31 days, February 29, March 31, April 30 May 31, June 30, July 31, August 31,

September 30, October 31, November 30, December 31. This averages out to!

We divide a circle into 360 units called degrees, which is six squared times ten. As such the

equilateral triangle has each angle equal to 60 degrees. There are approximately 360 earth

rotations in the time it orbits the sun once, Hence in one day the Earth moves through

approximately one degree in one day in its 365 day journey around the Sun. The sidereal

month, the time it takes the moon to return to the same position against the background of the

stars is 27 days 7 hours 43 minutes. We see the power of sexagesimal (base 60) in computing

time with the month divided into days, hours, and seconds where there are 60 minutes in an

1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 60

= 86,400

31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31

367

= 30.58333

of 54 57

hour and 60 seconds in a minute because we can convert this easily into seconds with such a

system as here, the duration of this sidereal month in seconds:!

The length of the synodic month, the completion of the phases of the moon, or the time it takes

the moon to return to the same position with respect to the Sun, is 29 days 12 hours 44

minutes 3 seconds. It is a little longer than the sidereal month because the earth has moved

with respect to the sun by the time the moon has come back around to a full orbit. We have:!

We looked at the arithmetic mean of the days in the month over a year, Let us look at the

harmonic mean:!

Gathering like terms in the denominator:!

=0.225806+0.034482759+0.133333=0.39362!

Which is the most frequent value for the days in a month. Now let’s look at the geometric mean

which tempers the various values with one another.!

=30.493!

But perhaps it is more telling to take the mean between the individual days of the month that

we use:!

0.032258+0.034482759+0.033333=0.10007!

(24 ⋅ 27 + 7) ⋅ 60

+ 43 ⋅ 60

+ 0 ⋅ 60

= 2,358,000 + 2,580 + 0 = 2360580sec /m onth

(24 ⋅ 29 + 12) ⋅ 60

+ 44 ⋅ 60

+ 3 ⋅ 60

= 2,548,800 + 2,640 + 3 = 2551443sec /month

0.032258 + 0.034482759 + 0.032258 + 0.033333 + 0.032258 + 0.033333 + 0.032258 + 0.032258 + 0.033333 + 0.032258 + 0.033333 + 0.032258

0.39362

= 30.486

31 ⋅ 29 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 30 ⋅ 31

31 + 29 + 30

= 30

0.10007

= 29.799 ≈ 30

31 ⋅ 29 ⋅ 30 = 29.98888 ≈ 30

of 55 57

We see that the arithmetic mean gives us exactly a 30 day month, hence we always speak

loosely saying a month is 30 days. !

Indeed when Buckminster Fuller speaks of Nature employing 60 degree coordination he is

speaking of the equilateral triangle as the fundamental unit of one for area. That is the six-sided

regular hexagon can be divided into six equilateral triangles, giving it an area of six. Indeed he

takes the equilateral triangle as one because not only does it enclose an area with the fewest

sides, it has the smallest area for its perimeter; a regular hexagon with a perimeter the same

size encloses more area. I would like to say the Ancient Sumerians, Babylonians, and Greeks

made this connection, because in their sexagesimal (base 60) system of notation, one is a

straight line with a triangle on top. See ﬁg 4. Ancient Mesopotamian Cuneiform for writing. It is

not sexagesimal notation in the strict sense, as they have a unique symbol for a tens column,

but it was this sexagesimal structure that lead to the 60 minute hour and 60 second minute, as

well as the angular division projected onto the sky in 60 minutes of arc in a degree and 60

seconds of arc in a minute as a way of mathematically deﬁning time dynamically in terms of the

motions of the Earth and moon.!

Fig. 4

of 56 57

14.0 The Cosmic Calendar

For equation 3.5 to be perfect Earth day needs to be shorter. A long time ago it was; the Earth

loses energy to the moon. The days become longer by 0.0067 hours per million years. Equation

3.5 is actually 1.2 seconds:

Equation 3.5

We have

24-20=0.0067t

t=597 million years

This was when the earth went through a dramatic change and there was a big explosion of life

(The Cambrian). The dinosaurs went extinct 65 million years ago giving small mammals a

chance to evolve paving the way for humans.

24-x=0.0067t

x=23.5645 hours

We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic

calendar:

What is the next term?

20+3+0.57735+0.4714=24hours

Which bring us to today.

h 4π r

≈ 6 ⋅

K Eof Moon

K Eof E ar th

⋅ Ear th Da y = (6)1.2secon d s

24h ours

1.2

= 20h ours

3cos(0

∘

) +

cos(30

∘

) = din osaur − ex t inct ion =

20hrs + 3hrs +

hrs +

3h ours +

h ours

of 57 57

The Author